Prove by minimum counterexample that $2^n>10n$ for $n>5$

Question

Prove by minimum counterexample that for all integers $n>5$ the statement $2^n>10n$ is true.

Attempt: Let $S$ be a set of counterexamples, $S=\{n \in \mathbb{Z_+}: 2^n \le 10n, \space n>5 \}$. Let $m \in S$ be the smallest element of $S$, so $m>5$. Then, $m-1 \notin S$. So, we have $$2^{m-1}>10(m-1) \\ 2^m>20(m-1)$$

I am stuck here. I dont know how to get rid of that $-1$. Help appreciated. Thank you.

Answer 1

Your attempt has some flaws. The method should show that there is no minimal counterexample by using the fact that eavery nonempty subset of $\mathbb Z_+$ has a minimal element. But you cannot show that $S$ is empty, simply because $S$ isn't empty (in fact $S=\{1,2,3,4,5\}$. So by actually taking the miniumal element of this $S$, you take $m=1$. Then indeed $m-1=0\notin S$, but so what? Adjust the definition of $S$ so that its emptyness is precisely the statement that there are no counterexamples to "$2^n>10n$ and $n>5$".